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主页 / user-274181

henriquehbr's questions

Martin Hope
henriquehbr
Asked: 2022-09-29 07:50:30 +0800 CST

无法通过 EFISTUB 直接启动 Linux 内核

  • 1

我已经从源代码编译了 Linux 内核 5.18.4,启用了所有与 EFI 相关的选项,没有任何内置参数,也没有默认的 init 路径,此外,值得一提的是,我没有使用 initramfs/initrd

我正在尝试在支持 EFI 的 VM 上通过 VirtualBox 6.1.34 启动此内核

安装盘(/dev/sda)有两个分区:

  • /dev/sda1格式化为 FAT32 的 512mb EFI 系统分区安装在/boot
  • /dev/sda2格式化为 ext4 的 15.5gb 根分区安装在/

内核位于/boot/EFI/BOOT/boox64.efi,似乎这个命名约定使它自动启动,跳过 UEFI shell,并且不需要通过创建启动条目efibootmgr,但我不确定这是否是不同硬件上不同 UEFI 实现的规范

每当我尝试启动它时,我都会收到以下错误:

虚拟机 UEFI 错误

在以前的尝试中,通过更改内核路径和进行其他一些调整等细微调整,结果是相似的,每当我尝试vmlinuz.efi通过 EFI shell 执行时,机器都会永远挂起,而不会显示一条错误消息(这是即使在传递root=/dev/sda2和init=/bin/init作为参数时也是如此)

linux kernel
  • 1 个回答
  • 25 Views
Martin Hope
henriquehbr
Asked: 2022-04-25 08:20:25 +0800 CST

使用内置的答案文件创建自定义 Alpine Linux ISO

  • 0

我找不到任何有关如何将文件添加到自定义 Alpine Linux ISO 的指南或文档,我能找到的最接近的是Alpine Wiki 上关于创建自定义 ISO 映像的页面mkimage

我宁愿将我的自动安装脚本和答案文件直接放在 ISO 上,而不必通过wget

alpine-linux iso
  • 2 个回答
  • 260 Views
Martin Hope
henriquehbr
Asked: 2022-03-31 08:25:49 +0800 CST

startx 不一致的行为

  • 1

我有一个 SSH 服务器,负责dwm通过 X 转发运行二进制文件,在我的客户端计算机上,我有一个 shell 脚本来替换上面的dwm二进制文件/usr/local/bin,在该脚本中,我只需调用服务器请求它运行原始文件dwm:

ssh -q -tt [email protected] dwm $@

只有一个-t,我得到以下错误:

Pseudo-terminal will not be allocated because stdin is not a terminal

在 SSH 服务器端,我正在使用ForceCommand通过container_runner脚本传递接收到的参数,它看起来像这样:

#!/bin/sh
/usr/local/bin/$SSH_ORIGINAL_COMMAND

这里的原因ForceCommand是为了限制运行除一组 Docker 容器初始化脚本(位于/usr/local/bin)之外的任何其他东西的可能性

这是其中一个脚本的示例,下面的一个运行dwmDocker 映像:

#!/bin/sh
docker run \
    --pull=never \
    --rm \
    -v container-scripts:/container-scripts \
    -v ssh-keys:/home/dwm/.ssh \
    -v x11-shared:/tmp/.X11-unix \
    -it \
    -e DISPLAY=:1 \
    -e XAUTHORITY=/tmp/.X11-unix/container-cookie \
    dwm \
    $@ 2>/dev/null
  • container-scripts- 包含一组脚本的卷,这些脚本将允许在容器内运行其他同级容器
  • ssh-keys- SSH 密钥,以便能够从其他容器中运行容器
  • x11-shared- X11 共享数据(X11 套接字和 xauth cookie)

我的ssh_config样子是这样的:

StrictHostKeyChecking no
UserKnownHostsFile /dev/null
ForwardX11 yes
ForwardX11Trusted yes
PreferredAuthentications=publickey

在我禁用 SSH 多路复用后,事情开始变得不一致ssh_config,之前使用下面的设置,它一直工作正常(这没有任何意义,据我所知,SSH 多路复用不应该干扰),下面的设置:

ControlPath /tmp/%r@%h:%p
ControlMaster auto
ControlPersist yes

发生的情况是,每当我尝试运行将我的脚本startx作为参数传递时,我的dwm脚本向 SSH 服务器发出请求以运行实际的dwm,我只是得到一个黑屏,并且没有任何反应,它只是保持这样,例如下面的示例:

startx /usr/local/bin/dwm -- :1

但是,如果我运行上面相同的命令,稍有不同,使用 shell 命令替换和 shell noop 运算符,它工作正常(只需要dwm比平时多几秒钟,可能是由于丑陋的黑客)

: $(startx /usr/local/bin/dwm -- :1)
shell-script ssh
  • 1 个回答
  • 140 Views
Martin Hope
henriquehbr
Asked: 2022-03-08 07:48:02 +0800 CST

SSH 服务器拒绝来自同一个客户端的多个连接

  • 1

我有一个在 Alpine Linux 3.15 Docker 容器中运行的简单 openssh 服务器,以及另一个容器,也运行相同版本的 Alpine 的 SSH 服务器)运行带有 dwm 的 X 服务器

我正在st使用以下命令从 X 服务器连接到 SSH 服务器(并正在运行):

ssh -v -t [email protected] st

SSH 服务器正在使用以下命令执行:

/usr/bin/sshd -D -d -e

$DISPLAY变量设置为ssh:10(作为sshSSH 服务器主机名)

客户端能够通过从服务器到客户端显示的 X11 转发打开图形应用程序( ),问题是在收到以下错误之前st我只能打开单个实例:st

ssh: connect to host 172.17.0.2 port 22: Connection refused

服务器端日志:

Starting session: command on pts/1 for client from 172.17.0.3 port 34060 id 0
debug1: Setting controlling tty using TIOCSCTTY.
debug1: X11 connection requested.
debug1: channel 3: new [X11 connection from 172.17.0.2 port 35028]
debug1: Received SIGCHLD.
debug1: session_by_pid: pid 9
debug1: session_exit_message: session 0 channel 0 pid 9
debug1: session_exit_message: release channel 0
debug1: session_by_tty: session 0 tty /dev/pts/1
debug1: session_pty_cleanup2: session 0 release /dev/pts/1
debug1: channel 3: free: X11 connection from 172.17.0.2 port 35028, nchannels 4
debug1: session_by_channel: session 0 channel 0
debug1: session_close_by_channel: channel 0 child 0
debug1: session_close_x11: detach x11 channel 1
debug1: session_close_x11: detach x11 channel 2
Close session: user client from 172.17.0.3 port 34060 id 0
debug1: channel 0: free: server-session, nchannels 3
debug1: channel 1: free: X11 inet listener, nchannels 2
debug1: channel 2: free: X11 inet listener, nchannels 1
Received disconnect from 172.17.0.3 port 34060:11: disconnected by user
Disconnected from user client 172.17.0.3 port 34060
debug1: do_cleanup
debug1: do_cleanup

客户端日志:

Running /usr/bin/xauth remove ssh:10.0
/usr/bin/xauth add ssh:10.0 MIT-MAGIC-COOKIE-1 49907415ff518044198f6f0075f270fe
debug1: client_input_channel_open: ctype x11 rchan 3 win 65536 max 16384
debug1: client_request_x11: request from 172.17.0.2 35028
debug1: channel 1: new [x11]
debug1: confirm x11
OpenSSH_8.8p1, OpenSSL 1.1.1l  24 Aug 2021
debug1: Reading configuration data /etc/ssh/ssh_config
debug1: Authenticator provider $SSH_SK_PROVIDER did not resolve; disabling
debug1: Connecting to 172.17.0.2 [172.17.0.2] port 22.
debug1: connect to address 172.17.0.2 port 22: Connection refused
ssh: connect to host 172.17.0.2 port 22: Connection refused
xinit: connection to X server lost

waiting for X server to shut down debug1: client_input_channel_req: channel 0 rtype exit-status reply 0
debug1: client_input_channel_req: channel 0 rtype [email protected] reply 0
debug1: channel 1: free: x11, nchannels 2
X connection to ssh:10.0 broken (explicit kill or server shutdown).
debug1: channel 0: free: client-session, nchannels 1
Connection to 172.17.0.2 closed.
Transferred: sent 91292, received 92728 bytes, in 3.1 seconds
Bytes per second: sent 29350.8, received 29812.5
debug1: Exit status 1
(II) Server terminated successfully (0). Closing log file.

deallocvt: can't open console
linux ssh
  • 1 个回答
  • 194 Views
Martin Hope
henriquehbr
Asked: 2022-02-27 07:10:22 +0800 CST

在 chroot 中安装期间内核版本不匹配

  • 0

我正在尝试通过我编写的自动化脚本在 QEMU 虚拟机上安装 Artix,在我设置 Rootless Docker 的最新安装步骤之一中,需要通过在安装挂载点内ip_tables运行来启用内核模块modprobe ip_tableschroot

问题是我收到以下错误:

modprobe: FATAL: Module ip_tables not found in directory `/lib/modules/5.13.4-artix1-1

经过进一步检查,我注意到pacman -Q linux返回版本5.16.10-artix1-1,而我的版本/lib/modules是5.13.4-artix1-1

跟这有关系吗chroot?还是问题出在其他地方?

arch-linux iptables
  • 1 个回答
  • 130 Views
Martin Hope
henriquehbr
Asked: 2019-12-03 03:52:37 +0800 CST

使用 systemctl 观察 systemd 服务状态

  • 10

是否有任何选项可以systemd实时查看服务状态systemctl

我实际上做什么:

systemctl status tor

例如:

systemctl watch tor
systemd manjaro
  • 1 个回答
  • 8335 Views
Martin Hope
henriquehbr
Asked: 2018-03-21 13:15:08 +0800 CST

dialog --menu 在 bash 上输出整数而不是字符串

  • 0

我在 bash 中创建了一个动态菜单,它从数组中获取它的项目,但是当用户选择特定项目时,菜单会显示返回码(0 或 1)

repositorios=() ; i=0
while read -r line; do
    let i=$i+1
repositorios+=($i "$line")
done < <( find ~ -type d -name .git )

gerenciar_repositorios=$(dialog --stdout --extra-button --help-button \
--ok-label "Acessar repositório" \
--extra-label 'Criar repositório' \
--cancel-label 'Remover repositório' \
--help-label 'Cancelar' \
--backtitle "Bem vindo ao Git Bash `whoami`!" \
--title ' Gerenciar repositórios ' \
--menu 'Gerenciar repositórios' 0 0 0 \
${repositorios[@]}) ; echo $gerenciar_repositorios

; echo $gerenciar_repositorios仅用于测试目的!

bash shell-script
  • 1 个回答
  • 1528 Views
Martin Hope
henriquehbr
Asked: 2018-03-21 04:59:05 +0800 CST

无法在 bash 上输出对话框值

  • 0

我在对话框上有一个动态菜单(这些项目是从数组中提供的),但菜单没有显示选定的选项

gp_options=()
for i in `find ~ -type d -name .git`; do
    gp_options+=("" "$i")
done

gp_dialog=(dialog --stdout --extra-button --help-button \
--ok-label 'Access repository' \
--extra-label 'Create repository' \
--cancel-label 'Remove repository'
--help-label 'Cancel' \
--backtitle "Welcome to Git Bash `whoami`!" \
--title ' Manage repositories ' \
--menu 'Manage repositories' \
0 0 0 \
"${gp_options[@]}")

dialog --stdout --msgbox "$manage_repositories" 0 0
bash shell-script
  • 1 个回答
  • 619 Views
Martin Hope
henriquehbr
Asked: 2018-03-05 08:20:41 +0800 CST

在对话框上显示变量--msgbox

  • 0

我想在对话框消息框上输出算术级数,但 --msgbox 属性将它们显示为纯文本

我的代码:

while :
do
    sequence=(1 10 100 1000 10000)
    dialog --stdout --msgbox '${sequence[*]}' 0 0
done
shell-script arithmetic
  • 1 个回答
  • 1170 Views
Martin Hope
henriquehbr
Asked: 2018-02-15 03:17:53 +0800 CST

Here-document 在 shell 脚本上无法正常工作

  • 1

我有两个文件:server.sh并且client.sh,服务器将以下文本发送到客户端

客户端.sh

porta_cliente=5678
porta_servidor=8765

nc -l $porta_cliente | sh

服务器.sh

porta_cliente=5678
porta_servidor=8765

nc $ip $porta_cliente <<-EOF
        dialog --stdout --backtitle 'Bem vindo ao SEPA 0.1' --title 'Menu inicial' --menu 'Escolha uma opção:' 0 0 0 \
        Cadastrar 'Criar um novo usuário' \
        Entrar 'Fazer login com sua conta' \
        Sair 'Encerrar o SEPA' > logservidor

        # WORKING FINE UNTIL HERE!

        case `cat logservidor` in
                "Cadastrar")
                        echo "LOG: case is being executed" > /dev/tcp/"$ip"/"$porta_servidor"
                        dialog --stdout --msgbox 'CADASTRAR' 0 0 > /dev/tcp/"$ip"/"$porta_servidor"
                ;;
                "Entrar")
                        echo "LOG: case is being executed"
                        echo "LOGIN SELECIONADO" > /dev/tcp/"$ip"/"$porta_servidor"
                ;;
                "Sair")
                        echo "LOG: case is being executed"
                        echo "SAIR SELECIONADO" > /dev/tcp/"$ip"/"$porta_servidor"
                ;;
        esac
EOF

如您所见,服务器通过 netcat 向客户端发送文本(解释为多行命令),在这些命令上,我试图将用户输入传递给特定文件,然后使用 case 语句检查用户输入是否Cadastrar,Entrar或者Sair,第一个对话框部分工作正常,但case 'cat log servidor' in没有被执行(即使bash -x server.sh我找不到导致这个问题的原因)

bash shell-script
  • 1 个回答
  • 2087 Views
Martin Hope
henriquehbr
Asked: 2018-02-14 04:16:11 +0800 CST

如何在 shell 脚本(netcat)上将文本作为代码从服务器执行到客户端

  • 1

我有两个文件,client.sh所有server.sh必要的数据都在服务器上,使用netcat发送到客户端,客户端只是获取这些数据并将其显示给最终用户,问题是当服务器发送一些多行代码时,客户只是将其作为文本接收并显示在屏幕上

客户端.sh

ip=127.0.0.1
client_port=5678
server_port=8765

while :
do
        # Request the menu from the server
        echo menu > /dev/tcp/"$ip"/"$server_port"
        # Waits the server response
        nc -l $porta_cliente
done

服务器.sh

ip=127.0.0.1
porta_cliente=5678
porta_servidor=8765

while :
do
        nc -vv -l $porta_servidor > logservidor
        echo "Texto recebido: "`cat logservidor` # LOG
        case `cat logservidor` in
        "splash")
                echo "dialog --stdout --msgbox 'SPLASH' 0 0" > /dev/tcp/"$ip"/"$porta_c$
        ;;
        "menu_inicial")
                nc $ip $porta_cliente <<-EOF
                        dialog --stdout --backtitle 'Bem vindo ao SEPA 0.1' --title 'Me$
                        Cadastrar 'Criar um novo usuário' \
                        Entrar 'Fazer login com sua conta' \
                        Sair 'Encerrar o SEPA'

                        # Caso o usuário selecione Cancelar, a execução do script será $
                        if [ $? -eq 0 ]; then
                                echo SUCESSO
                        else
                                rm resposta_servidor dados_digitados 2> /dev/null
                                clear
                                exit
                        fi
                EOF
        ;;
        "menu_principal")
                echo "dialog --msgbox 'MENU_PRINCIPAL' 0 0" > /dev/tcp/"$ip"/"$porta_cl$
        ;;
        *)
                dialog --msgbox 'WTF?!' 0 0 > /dev/tcp/"$ip"/"$porta_cliente"
        ;;
        esac
done
shell-script netcat
  • 2 个回答
  • 841 Views
Martin Hope
henriquehbr
Asked: 2018-02-12 09:07:20 +0800 CST

如何保持 netcat 连接打开?

  • 14

我有两个文件,client.sh和server.sh. 所有必要的数据都在服务器上,使用netcat发送到客户端。客户端只是获取这些数据并将其显示给最终用户。问题是,当我尝试将对话框加载屏幕从服务器显示到客户端时:

服务器.sh

# CLIENT PORT: 8765
# SERVER PORT: 5678

while :
do
    touch registered_users data

    nc -vv -l -p 5678 > data

    case `cat data` in
        "SPLASH_SCREEN")
            for ((i=0;i<100;i++))
            do
                echo $i
            done | dialog --title 'Loading...' --gauge 'Welcome!' 8 40 0 > /dev/tcp/127.0.0.1/8765
        ;;
    esac
done

客户端.sh

# CLIENT PORT: 8765
# SERVER PORT: 5678

echo "SPLASH_SCREEN" > /dev/tcp/127.0.0.1/5678

while :
do
    nc -l -p 8765 > server_response
    cat server_response
done
shell-script netcat
  • 1 个回答
  • 32146 Views
Martin Hope
henriquehbr
Asked: 2018-02-05 10:04:22 +0800 CST

案例不适用于 bash

  • 3

我已经修改了我的代码,但是我没有发现任何错误,当我尝试使用它执行时./client.sh会输出以下错误:

./client.sh: line 111: syntax error near unexpected token `done'

这是我的脚本:

while :
do
        clear
        menu_inicial=$(dialog --stdout --title 'Menu inicial' --menu 'Escolha uma opção:' 0 0 0 
        Cadastrar 'Cria um novo usuário'                            
        Entrar 'Fazer login com sua conta'                          
        Sair 'Encerra o SEPA')

        # Caso o usuário selecione Cancelar, a execução do script será encerrada
        if [ $? = 1 ]; then
            if [ -f resposta_servidor ] && [ -f dados_digitados ]; then
                rm resposta_servidor dados_digitados
                clear
                exit
            else
                clear
                exit
            fi
        fi

        # THE PROBLEM IS ON THIS CASE
        case "$menu_inicial" in

            "Entrar") ##### LOGIN #####
                while :
                do
                    clear
                    # Dialog para digitar o nome de usuário
                    login_usuario=$(dialog --title 'Entrar' --stdout --inputbox 'Digite seu nome de usuário:' 0 0)

                    # Caso o usuário selecione Cancelar, a execução do script será encerrada
                    if [ $? = 1 ]; then
                        break; # Retorna ao menu inicial
                    # Caso o nome de usuário nao for digitado, o usuário será alertado
                    elif [ "$login_usuario" = '' ]; then
                        dialog --stdout --msgbox 'Digite o nome de usuário!' 0 0
                        break;
                    fi

                    # Dialog para digitar a senha
                    login_senha=$(dialog --title 'Entrar' --stdout --passwordbox 'Digite sua senha:' 0 0)

                    # Caso o usuário selecione Cancelar, a execução do script será encerrada
                    if [ $? = 1 ]; then
                        break; # Retorna ao menu inicial
                    # Caso a senha não for digitada, o usuário será alertado
                    elif [ "$login_senha" = '' ]; then
                        dialog --stdout --msgbox 'Digite a senha!' 0 0
                        break;
                    fi

                    echo $login_usuario","$login_senha",login" > /dev/tcp/127.0.0.1/5678; # Envia os dados digitados pelo usuário para o servidor
                    nc -l -p 8765 > resposta_servidor # Abre a porta 8765 no cliente para receber a resposta do servidor

                    # Caso o nome de usuário de a senha estejam corretos, o login é feito
                    if grep -qw "LOGIN_SUCESSO" resposta_servidor
                    then
                        dialog --stdout --msgbox 'Conectado com sucesso!' 0 0
                        break;
                    # Caso o nome de usuário ou a senha estejam incorretos, uma mensagem de erro é mostrada
                    elif grep -qw "LOGIN_ERRO" resposta_servidor
                    then
                        dialog --stdout --msgbox 'Senha incorreta, tente novamente!' 0 0
                    fi
                done ##### END LOGIN #####
            ;;

            "Cadastrar") ##### REGISTER #####
                while :
                do
                    clear
                    # Dialog para digitar o nome de usuário
                    cadastrar_usuario=$(dialog --title 'Cadastrar' --stdout --inputbox 'Digite seu nome de usuário:' 0 0)

                    # Caso o usuário selecione Cancelar, a execução do script será encerrada
                    if [ $? = 1 ]; then
                        break; # Retorna ao menu inicial
                    # Caso o nome de usuário nao for digitado, o usuário será alertado
                    elif [ "$cadastrar_usuario" = '' ]; then
                        dialog --stdout --msgbox 'Digite o nome de usuário!' 0 0
                        break;
                    fi

                    # Dialog para digitar a senha
                    cadastrar_senha=$(dialog --title 'Cadastrar' --stdout --passwordbox 'Digite sua senha:' 0 0)

                    # Caso o usuário selecione Cancelar, a execução do script será encerrada
                    if [ $? = 1 ]; then
                        break; # Retorna ao menu inicial
                    # Caso a senha não for digitada, o usuário será alertado
                    elif [ "$cadastrar_senha" = '' ]; then
                        dialog --stdout --msgbox 'Digite a senha!' 0 0
                        break;
                    fi

                    echo $cadastrar_usuario","$cadastrar_senha",cadastro" > /dev/tcp/127.0.0.1/5678; # Envia os dados digitados pelo usuário para o servidor
                    nc -l -p 8765 > resposta_servidor # Abre a porta 8765 no cliente para receber a resposta do servidor

                    # Caso o nome de usuário de a senha estejam definidos, o cadastro é feito
                    if grep -qw "CADASTRO_SUCESSO" resposta_servidor
                    then
                        dialog --stdout --msgbox 'Cadastrado com sucesso!' 0 0
                        break;
                    # Caso ocorra algum erro durante o cadastro, uma mensagem de erro é mostrada
                    elif grep -qw "CADASTRO_ERRO" resposta_servidor
                    then
                        dialog --stdout --msgbox 'Erro no cadastro, tente novamente!' 0 0
                    fi
                done ##### END REGISTER #####
            ;;
        esac
done
shell-script
  • 1 个回答
  • 1294 Views
Martin Hope
henriquehbr
Asked: 2018-02-05 08:16:16 +0800 CST

如何在 shell 脚本中比较两个文件的字符串?

  • 2

第一个字段是用户名,第二个是密码,最后一个表示操作(登录/注册)

文件 1:用户输入

hello,world,register

文件 2:纯文本数据库

admin,123
user,321
foo,bar

如何将用户输入与数据库进行比较以检查用户名是否已存在?

我已经尝试了以下代码,但它不起作用:

if cut -d "," -f1 user_input | grep -wf user_input database; then
    echo "This username is already in use, try again!"
else
    echo "Successfully registered!"
fi

PS:我需要一个没有 awk、sed 或 find 的解决方案,只需 grep 和 cut

shell-script grep
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